Problem: Simplify and expand the following expression: $ \dfrac{p - 8}{p + 4}+\dfrac{p + 2}{3p + 9} $
Solution: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(p + 4)(3p + 9)$ Multiply the first term by $\dfrac{3p + 9}{3p + 9}$ $ \begin{align*} \dfrac{p - 8}{p + 4} \times \dfrac{3p + 9}{3p + 9} & = \dfrac{(p - 8)(3p + 9)}{(p + 4)(3p + 9)} \\ & = \dfrac{3p^2 - 15p - 72}{(p + 4)(3p + 9)}\end{align*} $ Multiply the second term by $\dfrac{p + 4}{p + 4}$ $ \begin{align*} \dfrac{p + 2}{3p + 9} \times \dfrac{p + 4}{p + 4} & = \dfrac{(p + 2)(p + 4)}{(3p + 9)(p + 4)} \\ & = \dfrac{p^2 + 6p + 8}{(3p + 9)(p + 4)}\end{align*} $ Now we have: $ = \dfrac{3p^2 - 15p - 72}{(p + 4)(3p + 9)} + \dfrac{p^2 + 6p + 8}{(3p + 9)(p + 4)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{3p^2 - 15p - 72 + p^2 + 6p + 8}{(p + 4)(3p + 9)} $ $ = \dfrac{4p^2 - 9p - 64}{(p + 4)(3p + 9)}$ Expand the denominator: $ = \dfrac{4p^2 - 9p - 64}{3p^2 + 21p + 36}$